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- == Problem == [[Image:AIME_1998-6.png|350px]]2 KB (254 words) - 19:38, 4 July 2013
- == Problem == If <math>1998</math> is written as a product of two positive integers whose difference is1 KB (157 words) - 14:28, 5 July 2013
- ==Problem== <math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>2 KB (294 words) - 22:14, 17 December 2024
- == Problem == {{USAMO newbox|year=1998|num-b=5|after=Last Question}}5 KB (871 words) - 18:59, 10 May 2023
- ==Problem== Determine the least possible value of <math>f(1998),</math> where <math>f:\Bbb{N}\to \Bbb{N}</math> is a function such that fo454 bytes (73 words) - 01:54, 28 August 2024
- == Problem ==1 KB (138 words) - 17:30, 12 May 2024
- == Problem == Find the remainder of the division of <math>x_{1998}</math> by <math>1998</math>.720 bytes (117 words) - 15:57, 13 December 2023
Page text matches
- ...cluding Art of Problem Solving, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete be ...ics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.11 KB (1,517 words) - 14:11, 2 March 2025
- This is a problem where constructive counting is not the simplest way to proceed. This next e ...wer is <math>8 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 = 8 \cdot 7^6 = 941,192</math>, as desired. <math>\square</math>13 KB (2,018 words) - 15:31, 10 January 2025
- ...iv style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div> ...? <div style="text-align:right">([[1998 AIME Problems/Problem 8|1998 AIME, Problem 8]])</div>7 KB (1,111 words) - 14:57, 24 June 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1999 AIME Problems/Problem 1|Problem 1]]1 KB (118 words) - 08:41, 7 September 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1998 AIME Problems]]1 KB (114 words) - 08:39, 7 September 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1997 AIME Problems/Problem 1|Problem 1]]1 KB (114 words) - 08:39, 7 September 2011
- == Problem 1 == [[1997 AIME Problems/Problem 1|Solution]]7 KB (1,098 words) - 17:08, 25 June 2020
- {{AIME Problems|year=1998}} == Problem 1 ==7 KB (1,084 words) - 02:01, 28 November 2023
- == Problem 1 == [[1999 AIME Problems/Problem 1|Solution]]7 KB (1,094 words) - 13:39, 16 August 2020
- == Problem == ...r this point it must continue to repeat. Thus, in particular <math>a_{j + 6} = a_j</math> for all <math>j</math>, and so repeating this <math>n</math>2 KB (406 words) - 11:20, 15 February 2025
- == Problem == { }_{2} X_{4},{ }_{4} X_{6}, \ldots,{ }_{n-2} X_{n},(n, n+1)(n+1,1)(1, n+2)(n+2,2)9 KB (1,659 words) - 18:35, 20 June 2024
- == Problem == | 0 || 1 || 2 || 3 || 4 || 5 || 62 KB (354 words) - 13:19, 14 December 2024
- == Problem == Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain5 KB (684 words) - 18:52, 19 June 2024
- == Problem == [[Image:AIME_1998-6.png|350px]]2 KB (254 words) - 19:38, 4 July 2013
- == Problem == Though the problem may appear to be quite daunting, it is actually not that difficult. <math>\1 KB (225 words) - 16:56, 3 February 2025
- == Problem == ...m{2}{2} = 6</math> ways. This gives us a total of <math>10 \cdot 2 \cdot 6 = 120</math> possibilities in which all three people get odd sums.5 KB (917 words) - 02:37, 12 December 2022
- == Problem == ...count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the d6 KB (913 words) - 16:34, 6 August 2020
- == Problem == ...{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math>, <math>8^8</math>, and <math>k</math>?2 KB (289 words) - 22:50, 23 April 2024
- ...tts Institute of Technology]] (MIT) each winter. It has been running since 1998. The contest takes place alternatively at Harvard or MIT each year. It is c ...of|difficulty=5-8|breakdown=<u>Individual</u>: 5 (Problem 1-5), 6 (Problem 6-10)<br><u>Team</u>: 7.5<br><u>HMIC</u>: 8}}4 KB (539 words) - 16:58, 19 February 2023
- * [[1959 IMO Problems/Problem 1 | Problem 1]] proposed by Poland * [[1959 IMO Problems/Problem 2 | Problem 2]] proposed by Constantin Ionescu-Tiu, Romania36 KB (4,083 words) - 21:48, 8 December 2024
